# The Ascherman method

The Ascherman method is alternative method of matchpoint scoring in pairs contests. We show here it's close relation to Neuberg. In fact it renders the Neuberg formula redundant.

### Percentage according to Ascherman

As we saw before, in the calculation of the result for hands with a different number of scores we may use the Neuberg formula. We will now look at a different, closely related, method, the Ascherman method. For the conversion of percentages from n to N scores according to Neuberg we have:
 PN = (Pn − 50) · n−1 n · N N −1 + 50
(1)
In the Ascherman method we use instead:
 PA = (Pn − 50) · n−1 n + 50
(2)
The nice thing about this percentage according to Ascherman is that it does not matter what we choose for N. N does not appear in the formula.
Our example in the derivation of Neuberg:

 session 1        3 tables session 2        6 tables score number MP percent score number MP percent 430 1 4 100 430 2 9 90 400 1 2 50 400 2 5 50 ‑50 1 0 0 ‑50 2 1 10
becomes with this version of the percentage

 session 1        3 tables session 2        6 tables score number MP percent score number MP percent 430 1 4 83.3 430 2 9 83.3 400 1 2 50 400 2 5 50 -50 1 0 16.7 -50 2 1 16.7
We see two things happening:
1. The percentages are independent on the number of scores.
2. The percentages are closer to 50%.
The first point is of course the most interesting aspect. As a result, for hands with a different tops, scores expressed in percentages are directly comparable. We don't have to do anything when converting to a different number of scores. Neuberg becomes reduntant.
When N goes to infinity the formulae (1) and (2) are equal.
Ascherman is the limit for N → ∞ of Neuberg.
The second point requires some getting used to. The range of possible percentages is no longer {0 ... 100} but {50/n ... 100 − 50/n} . Only if the number of scores is infinitely large 0 and 100 are obtainable.

### Matchpoints

In the classical method of scoring we use matchpoints S = 0, 2, ... 2n−2. The percentage scores are calculated using
 Pn = 100 · S 2n−2
(3)
In order to convert this to Ascherman percentages we have, with (2)
 PA = (100 · S 2n−2 −50 ) · n−1 n + 50
(4)
which may be reduced to
 PA = 100 · S + 1 2n
(5)
In the Ascherman method it is customary to use different matchpoints, running from 1:
SA = 1, 3, ... 2n−1. In other words
 SA = S+1
(6)
We see from (5) and (6) that
 PA = 100 · SA 2n
(7)
Notice that to calculate the percentage according ot Ascherman we might use (5) as well (7). Introducing the Ascherman matchpoints (6) therefore is not essential for the Ascherman method.

As we saw above it is not necessary to modify the percentage score when converting from one group size to another. For the score in matchpoints this is not true. The equivalent of the Neuberg formula (equation (10) on the Neuberg page) is

 SNA = SnA · N n
(8)
or, one determines the Ascherman matchpoints in the separate group with n scores and multiplies the result by N/n. This is not too complicated, even when scoring manually.

### A different view at Ascherman

We found Ascherman as a limiting case of Neuberg. Using the directives from Law 78A we may arrive at the Ascherman method in a direct way. Assume on a board with a total of n scores a certain score occurs m times, while there are k lower scores, then the number of matchpoints corresponding to this score is given by
 S = 2k + m - 1
(9)
In the classical method we obtain with equation (3) a percentage of
 Pn = 100 · 2k + m - 1 2n−2
(10)
One notices immediately this does not scale nicely with the size of the field. That is to say if each of k, m and n are multiplied by the same factor we would like P to remain constant. An obvious way to reach this goal is to drop the -1 and -2 from the formula, and use instead:
 PA = 100 · 2k + m 2n
(11)
and there we are, we have obtained Ascherman! See equation(5).

In the Ascherman method the numbers ak, am, an yield exactly the same score as the numbers k, m, n:

 PA(ak,am,an)  =  PA(k,m,n)
(12)

An example: Consider a board where always precisely half the participants go 1 down (the others make the contract). With the classical method the score will be 0 if there are 2 pairs, 16.67% when there 4 pairs, 20% for 6 pairs, 22.22% for 10 pairs, ...
With the Ascherman method the score is 25% in all these cases.

The classical method does not have this neat property. However, one easily verifies that substitution of (ak, am, an) for (k,m,n) in equation (10), with a=N/n, yields the Neuberg formula.

### A variation used in Norway

Sven Pran describes on BLML a method used in Norway.
Instead of (9) the matchpoints are defined by
 SN= 2k + m - n
(13)

such that the matchpoints are symmetrically around 0 and run from -(n - 1) to +(n - 1).

If this is divided by n we get a measure of the performance that is independent of N, as in the original Ascherman method.
Again, eq (8) applies.

http://tameware.com/adam/bridge/laws/blml/blml_archives/2014-April.txt (go to Apr 16 20:18:31 2014)

### History

The Dutchman W. Ascherman introduced this method in 1949 in the periodical "Bridge", as "a generally valid and correct method of assessment, independent of the value of n". In the same article, as an aside, the Neuberg formula, that Neuberg rediscovered many years later, was already given. Ascherman added "Thus it is desirable that the current regulation is replaced as soon as possible". The Dutch regulating authorities ignored this demand. In Belgium the Ascherman method did find a strong promotor in Herman De Wael. Without his work Ascherman would probably have been completely been forgotten.

Formulae translated from TEX by TTH, version 4.03.